In the circuit shown, the phase currents are
\[
I_A=572.812+j50.115\text{ A}
\]
\[
I_B=-254.525-j459.175\text{ A}
\]
\[
I_C=-207.083+j444.091\text{ A}
\]
Given that the CTs are ideal with no saturation, and the turns ratio of the Main CT is \(300:5\) and that of the Auxiliary Transformer (\(Yn\Delta\)) is \(2:1\) on every phase, the value of \(I_{AR}\), rounded off to three decimal places, is:
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First scale the phase currents down through the 300:5 Main CT, then subtract the zero sequence average (IA+IB+IC)/3 from the phase A current to get IAR.
Step 1: Work with the raw phase currents first.
Add the three given currents directly:
\[ I_A+I_B+I_C=(572.812-254.525-207.083)+j(50.115-459.175+444.091)=111.204+j35.031 \]
Step 2: Get the zero sequence part on the primary side.
\[ I_{0}=\frac{111.204+j35.031}{3}=37.068+j11.677 \]
Step 3: Subtract it from phase A on the primary side.
Since $I_{AR}$ is built from phase A with the zero sequence content removed, on the primary side this quantity is
\[ I_A-I_{0}=(572.812-37.068)+j(50.115-11.677)=535.744+j38.438 \]
\[ |I_A-I_0|=\sqrt{535.744^2+38.438^2}=537.24\text{ A},\qquad \angle=\tan^{-1}\left(\frac{38.438}{535.744}\right)=4.105^{\circ} \]
This matches option (C) exactly, but option (C) is the value before the Main CT scales it down, so it cannot be the final relay current.
Step 4: Apply the Main CT ratio once, at the end.
Only now divide by the Main CT ratio of $60$, since $300:5=60:1$, because the ratio is linear and can be applied after combining the currents instead of before:
\[ I_{AR}=\frac{537.24\angle4.105^{\circ}}{60}=8.954\angle4.105^{\circ}\text{ A} \]
Step 5: Confirm the angle is unaffected.
Dividing by the real number $60$ scales the magnitude only and leaves the angle at $4.105^{\circ}$, exactly matching option (D).
\[ \boxed{8.954\angle4.105^{\circ}\text{ A}} \]
